rút gọn biểu thức
\(A=\frac{8\sqrt{41}}{\sqrt{45+4\sqrt{41}}+\sqrt{45-4\sqrt{41}}}:\left(\sqrt{3}-\sqrt{2}\right)\)
mọi người giải đầy đủ giúp em với ạ
Mọi người giúp mình mấy câu rút gọn này với
A= \(\frac{8\sqrt{41}}{\sqrt{45+4\sqrt{41}}+\sqrt{45+4\sqrt{41}}}\)
B= \(\left(\frac{2x+1}{x\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\right)\div\left(1-\frac{x+4}{x+\sqrt{x}+1}\right)\)
A = \(\frac{8\sqrt{41}}{2\sqrt{2^2+2.2.\sqrt{41}+\sqrt{41}^2}}\)
A = \(\frac{8\sqrt{41}}{2\sqrt{\left(2+\sqrt{41}\right)^2}}\)
A = \(\frac{8\sqrt{41}}{2\left|2+\sqrt{41}\right|}\)
A = \(\frac{8\sqrt{41}}{4+2\sqrt{41}}\)
B = \(\left(\frac{2x+1}{\sqrt{x}^3+1^3}-\frac{1}{\sqrt{x}-1}\right):\frac{x+\sqrt{x}+1+x+4}{x+\sqrt{x}+1}\)
B = \(\left(\frac{2x+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}-1}\right).\frac{x+\sqrt{x}+1}{2x+\sqrt{x}+5}\)
Bạn tự làm tiếp nhé, mỏi tay quá!!
\(A=\frac{8\sqrt{41}}{2\sqrt{45+4\sqrt{41}}}=\frac{8\sqrt{41}}{2\sqrt{41+4\sqrt{41}+4}}=\frac{8\sqrt{41}}{2\sqrt{\left(\sqrt{41}\right)^2+2\cdot\sqrt{41}\cdot2+2^2}}\)
\(=\frac{8\sqrt{41}}{2\sqrt{\left(\sqrt{41}+2\right)^2}}=\frac{8\sqrt{41}}{2\left(\sqrt{41}+2\right)}=\frac{8\sqrt{41}\left(\sqrt{41}-2\right)}{2\left(41-4\right)}=\frac{328-16\sqrt{41}}{74}=\frac{164-8\sqrt{41}}{37}\)
\(B=\left(\frac{2x+1}{x\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\right):\left(1-\frac{x+4}{x+\sqrt{x}+1}\right)\)
\(=\left(\frac{2x+1}{\sqrt{x}^3+1^3}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right):\left(\frac{x+\sqrt{x}+1-x-4}{x+\sqrt{x}+1}\right)\)
\(=\left(\frac{2x+1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right):\left(\frac{\sqrt{x}-3}{x+\sqrt{x}+1}\right)\)
\(=\frac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\frac{x+\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\frac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}}{\sqrt{x}-3}=\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}=\frac{x+3\sqrt{x}}{x-9}\)
Rút gọn biểu thức :
\(M=\frac{8\sqrt{41}}{\sqrt{45+4\sqrt{41}+\sqrt{45}-4\sqrt{41}}}\)
Dấu căn dưới mẫu có vt lộn ko z bạn?
à nó bị liền : \(M=\frac{8\sqrt{41}}{\sqrt{45+4\sqrt{41}}+\sqrt{45-4\sqrt{41}}}\)
Đề sau khi đã sửa: \(M=\frac{8\sqrt{41}}{\sqrt{45+4\sqrt{41}}+\sqrt{45-4\sqrt{41}}}\)
Giải:
\(M=\frac{8\sqrt{41}}{\sqrt{\left(\sqrt{41}+2\right)^2}+\sqrt{\left(\sqrt{41}-2\right)^2}}\)
\(M=\frac{8\sqrt{41}}{\sqrt{41}+2+\sqrt{41}-2}\) = \(\frac{8\sqrt{41}}{2\sqrt{41}}=4\)
Vậy M = 4
\(\frac{8\sqrt{41}}{\sqrt{45+4\sqrt{41}}+\sqrt{45-4\sqrt{41}}}:\left(\sqrt{3}-\sqrt{2}\right)\)
\(\frac{8\sqrt{41}}{\sqrt{45+4\sqrt{41}+\sqrt{45-\sqrt{41}}}}:\left(\sqrt{3}-\sqrt{2}\right)\) ( đề)
\(=\frac{8\sqrt{41}}{\sqrt{41}+2-\sqrt{41}-2}:\left(\sqrt{3}-\sqrt{2}\right)\)
\(=2\sqrt{41}:\left(\sqrt{3}-\sqrt{2}\right)\)
\(=2\sqrt{123}+2\sqrt{82}\)
vậy.....................
Rút gọn:
\(A=\dfrac{8\sqrt{41}}{\sqrt{45+4\sqrt{41}}+\sqrt{45-4\sqrt{41}}}:\left(\sqrt{3}-\sqrt{2}\right)\)
\(A=\dfrac{8\sqrt{41}}{\sqrt{41}+2+\sqrt{41}-2}\cdot\dfrac{1}{\sqrt{3}-\sqrt{2}}\)
\(=\dfrac{4}{\sqrt{3}-\sqrt{2}}=4\sqrt{3}+4\sqrt{2}\)
Tính
\(\frac{8\sqrt{41}}{\sqrt{45+4\sqrt{41}}+\sqrt{45-4\sqrt{41}}}\)
Help me !!!! ~~ Giúp em với ạ T.T <3 Tks nhìu <3 <3
\(\frac{8\sqrt{41}}{\sqrt{45+4\sqrt{41}}+\sqrt{45-4\sqrt{41}}}\)
\(=\frac{8\sqrt{41}}{\sqrt{41+2.2.\sqrt{41}+4}+\sqrt{41-2.2.\sqrt{4}+4}}\)
\(=\frac{8\sqrt{41}}{\sqrt{\left(\sqrt{41}+2\right)^2}+\sqrt{\left(\sqrt{41}-2\right)^2}}\)
\(=\frac{8\sqrt{41}}{\sqrt{41}+2+\sqrt{41}-2}\)
\(=\frac{8\sqrt{41}}{2\sqrt{41}}=4\)
\(\dfrac{8\sqrt{41}}{\left(\sqrt{45}+4\sqrt{41}\right)\left(\sqrt{45}-4\sqrt{41}\right)}\)
Rút gọn
M=\(\dfrac{8\sqrt{41}}{\sqrt{45+4\sqrt{41}}+\sqrt{45-4\sqrt{41}}}\)
\(M=\dfrac{8\sqrt{41}}{\sqrt{45+4\sqrt{41}}+\sqrt{45-4\sqrt{41}}}\)
\(M=\dfrac{8\sqrt{41}}{\sqrt{\left(\sqrt{41}+2\right)^2}+\sqrt{\left(\sqrt{41}-2\right)^2}}\)
\(M=\dfrac{8\sqrt{41}}{\sqrt{41}+2+\sqrt{41}-2}\)
\(M=\dfrac{8\sqrt{41}}{2\sqrt{41}}=\dfrac{8}{2}=4\)
Vậy M = 4
Học tốt nhé :)
Rút gọn biểu thức
a) \(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{45}+\dfrac{5-\sqrt{5}}{\sqrt{5}}\)
b) \(\sqrt{7-4\sqrt{3}}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
giúp em với ạ, em cảm ơn!
a) \(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{45}+\dfrac{5-\sqrt{5}}{\sqrt{5}}=\sqrt{5}+\sqrt{5}+\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}}=\sqrt{5}+\sqrt{5}+\sqrt{5}-1=-1+3\sqrt{5}\)
b) \(\sqrt{7-4\sqrt{3}}+\sqrt{\left(1+\sqrt{3}\right)^2}=\sqrt{\left(2-\sqrt{3}\right)^2}+1+\sqrt{3}=2-\sqrt{3}+1+\sqrt{3}=3\)
a: \(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{45}+\dfrac{5-\sqrt{5}}{\sqrt{5}}\)
\(=\sqrt{5}+\sqrt{5}+\sqrt{5}-1\)
\(=3\sqrt{5}-1\)
b: \(\sqrt{7-4\sqrt{3}}+\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=2-\sqrt{3}+\sqrt{3}+1\)
=3
tìm x thoả mãn: \(x^3+\frac{8\sqrt{41}}{\sqrt{45+4\sqrt{41}}+\sqrt{45-4\sqrt{41}}}x+5=0\)
Xét \(\frac{8\sqrt{41}}{\sqrt{45+4\sqrt{41}}+\sqrt{45-4\sqrt{41}}}=\frac{8\sqrt{41}}{\sqrt{\left(\sqrt{41}+2\right)^2}+\sqrt{\left(\sqrt{41}-2\right)^2}}=\frac{8\sqrt{41}}{\sqrt{41}+2+\sqrt{41}-2}=\frac{8\sqrt{41}}{2\sqrt{41}}=4\)
Phương trình trên tương đương:
x3+4x+5=0
<=>x(x2-1)+5(x+1)=0
<=>x(x-1)(x+1)+5(x+1)=0
<=>(x+1)(x2-x+5)=0
<=>x+1=0 hoặc x2-x+5=0(vô nghiệm)
<=>x=-1
Vậy pt trên có nghiệm là x=-1
Bài này đi thi vio mk cũng gặp ..
bằng 1 ak